111 lines
3.1 KiB
Java
111 lines
3.1 KiB
Java
package ds.tree;
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import leecode.tree.TreeNode;
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import java.util.Stack;
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/**
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* @author Wen
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* @date 2022/9/1 22:03
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*/
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public class TreeTraverse {
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//前序
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public void pre(TreeNode h) {
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if (h == null) {
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return;
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}
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Stack<TreeNode> stack = new Stack<>();
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stack.push(h);
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while (!stack.isEmpty()) {
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TreeNode pop = stack.pop();
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System.out.println(pop.val);
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if (pop.right != null) {
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stack.push(pop.right);
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}
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if (pop.left != null) {
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stack.push(pop.left);
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}
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}
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}
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//中序
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public void in(TreeNode h) {
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if (h == null) {
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return;
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}
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Stack<TreeNode> stack = new Stack<>();
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while (!stack.isEmpty() || h != null) {
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if (h != null) {
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stack.push(h);
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h = h.left;
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} else {
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h = stack.pop();
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System.out.println(h.val);
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h = h.right;
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}
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}
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}
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//后序
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public void pos(TreeNode h) {
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if (h == null) {
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return;
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}
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Stack<TreeNode> stack = new Stack<>();
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Stack<TreeNode> stack2 = new Stack<>();
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stack.push(h);
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while (!stack.isEmpty()) {
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TreeNode pop = stack.pop();
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stack2.push(pop);
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if (pop.left != null) {
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stack.push(pop.left);
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}
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if (pop.right != null) {
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stack.push(pop.right);
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}
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}
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while (!stack2.isEmpty()) {
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System.out.println(stack2.pop().val);
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}
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}
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/**
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* c标记树的处理路径，先处理左树。
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* 没有左树时开始弹出，弹出后h移动到c的位置记录上一次的弹出位置
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* c指向下一个将要弹出的节点，这时候将判断h与c的位置关系，
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* h如果既不是c的左子树也不是右子树，按后序遍历的顺序来看相当于c还未处理左右子树，此时优先处理左子树，然后处理右子树
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*/
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public void posPro(TreeNode h) {
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if (h == null){
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return;
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}
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TreeNode c;
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Stack<TreeNode> stack = new Stack<>();
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stack.push(h);
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while (!stack.isEmpty()) {
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//指向将要弹出的节点，但不弹出
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c = stack.peek();
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if (c.left != null && h != c.left && h != c.right) {
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//c还没处理左子树
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stack.push(c.left);
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c = c.left;
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} else if (c.right != null && h != c.right) {
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//c还没处理右子树
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stack.push(c.right);
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c = c.right;
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} else {
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//c可以弹出了
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c = stack.pop();
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System.out.println(c.val);
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h = c;
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}
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}
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}
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public static void main(String[] args) {
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TreeTraverse treeTraverse = new TreeTraverse();
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treeTraverse.posPro(TreeNode.parseTree(new Integer[]{1,2,3,4,5,6,7}));
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}
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}
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