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[Algorithm] 94. Binary Tree Inorder Traversal iteratively approach

时间:2019-12-05 22:48:52      阅读:215      评论:0      收藏:0      [点我收藏+]

标签:while   turn   when   track   first   put   class   ber   null   

Given a binary tree, return the inorder traversal of its nodes‘ values.

Example:

Input: [1,null,2,3]
   1
         2
    /
   3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */

function getLefts(root) {
    if (!root) {
        return [];
    }
    
    let result = [];
    let current = root;
    while(current) {
        result.push(current);
        current = current.left ;
    }
    
    return result;
}

/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var inorderTraversal = function(root) {
    let lefts = getLefts(root);
    let result = [];
    while (lefts.length) {
        let newRoot = lefts.pop();
        result.push(newRoot.val);
        if (newRoot.right) {
            let newLefts = getLefts(newRoot.right);
            lefts = lefts.concat(newLefts)
        }
    }
    
    return result;
};

The idea is 

  • get all the left side node and push into stack
  • because stack is first in last out, when we do pop(), we are actually start from bottom of the tree.
  • everytime, we pop one node from stack, we check its right child, get all the left nodes from this right child and append into stack.
  • in this way, we are able to track all the node.

[Algorithm] 94. Binary Tree Inorder Traversal iteratively approach

标签:while   turn   when   track   first   put   class   ber   null   

原文地址:https://www.cnblogs.com/Answer1215/p/11992069.html

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