算法练习day15

一、平衡二叉树

递归法，后序

高度只能从下往上找，所以必须用后序遍历，左右中

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isBalanced = function (root) {
    function getDepth(root) {n
        if (!root) {
            return 0
        }
        let leftHeight = getDepth(root.left)
        let rightHeight = getDepth(root.right)
        if (leftHeight === -1 || rightHeight === -1 || Math.abs(leftHeight - rightHeight) > 1) {
            return -1
        } 
        return Math.max(leftHeight, rightHeight) + 1
    }
    return getDepth(root) !== -1
};

二、二叉树的所有路径

求根节点到叶子节点的路径，所以需要前序遍历

回溯思路，记录的路径需要回退再进入另一个路径，递归和回溯是相伴相生的，有一次递归就需要一次回溯

递归法

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {string[]}
 */
var binaryTreePaths = function(root) {
    let result = []
    function getPath(node, curPath = '') {
        if(!node.left && !node.right) {
            result.push(curPath + node.val)
            return
        }
        curPath += node.val + '->'
        node.left && getPath(node.left, curPath)
        node.right && getPath(node.right, curPath)
    }
    getPath(root)
    return result
};

迭代法

var binaryTreePaths = function(root) {
    let result = []
    let paths = ['']
    let stack = [root]

    while(stack.length) {
        let node = stack.pop()
        let path = paths.pop()
        if(!node.left && !node.right) {
            result.push(path + node.val)
            continue
        }
        path += node.val + '->'
        if(node.right) {
            paths.push(path)
            stack.push(node.right)
        }
        if(node.left) {
            paths.push(path)
            stack.push(node.left)
        }
    }

    return result
};

三、左叶子之和

左叶子节点的定义：节点A的左节点不为空，且左节点的左右子节点都为控，叶子节点

递归法

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var sumOfLeftLeaves = function(root) {
    function getSum(node) {
        if(!node) {
            return 0
        }
        let midVal = 0
        if(node.left && !node.left.left && !node.left.right) {
            midVal += node.left.val
        }
        let leftVal = getSum(node.left)
        let rightVal = getSum(node.right)
        return midVal + leftVal + rightVal
    }
    return getSum(root)
};

迭代法

var sumOfLeftLeaves = function(root) {
    let stack = [root]
    let sum = 0
    while(stack.length) {
        let node = stack.pop()
        if(node.left && !node.left.left && !node.left.right) {
            sum += node.left.val
        }
        node.right && stack.push(node.right)
        node.left && stack.push(node.left)
    }
    return sum
};